Let $h$ be a vector-valued function defined by $h(t)=(\log(t+100),\sin(3t))$. Find $h$ 's second derivative $h''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-\dfrac{1}{(t+100)^2},-9\sin(3t)\right)$ (Choice B) B $\left(\dfrac{1}{\ln(10)(t+100)},3\cos(3t)\right)$ (Choice C) C $\left(\dfrac{1}{(t+100)^2},-\cos(3t)\right)$ (Choice D) D $\left(-\dfrac{1}{\ln(10)(t+100)^2},-9\sin(3t)\right)$
Answer: We are asked to find the second derivative of $h$. This means we need to differentiate $h$ twice. In other words, we differentiate $h$ once to find $h'$, and then differentiate $h'$ (which is a vector-valued function as well) to find $h''$. Recall that $h(t)=(\log(t+100),\sin(3t))$. Therefore, $h'(t)=\left(\dfrac{1}{\ln(10)(t+100)},3\cos(3t)\right)$. Now let's differentiate $h'(t)=\left(\dfrac{1}{\ln(10)(t+100)},3\cos(3t)\right)$ to find $h''$. $h''(t)=\left(-\dfrac{1}{\ln(10)(t+100)^2},-9\sin(3t)\right)$ In conclusion, $h''(t)=\left(-\dfrac{1}{\ln(10)(t+100)^2},-9\sin(3t)\right)$.